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In proving the theorem: Every infinite set is equivalent to one of its proper subsets, I am confused about the following:

We consider an infinite set $M$, which always contains a countable subset, which is denoted $A := \{a_1, a_2, \dots \}$. We may partition $A$ into two countable subsets:

$$ A_1 := \{a_1, a_3, a_5 \dots \}, \qquad A_2 := \{a_2, a_4, a_6 \dots \} $$

and we have a one-to-one correspondence between $A$ and $A_1$ given by $a_n \to a_{2n-1}$.

We can then extend this correspondence to a one-to-one correspondence between the two sets:

$$ A \cup (M - A) = M, \qquad A_1 \cup (M-A) = M-A_2 $$

by simply assigning $x$ itself to each $x \in M-A$. I don't quite understand this extension, how are we allowed to do this?

This is a proof taken from Introductory Real Analysis by Kolmogorov and Fomin.

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  • $\begingroup$ What do you mean by "being allowed"? It is not a matter of permissions. $\endgroup$ – Andrés E. Caicedo Feb 17 '18 at 3:02
  • $\begingroup$ @AndrésE.Caicedo as in what is the property that permits us to perform this extension $\endgroup$ – dimebucker Feb 17 '18 at 3:04
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You have a bijection between the elements of $A$ and $A_{1}$. That is, you have a map $f\colon A\longrightarrow A_{1}$ which is bijective. You define another map, $g\colon A\cup (M-A) \longrightarrow A_{1}\cup (M-A)$, such that $$g(x)= f(x) \quad \text{if} \quad x\in A,\quad \text{and} \quad g(x)=x \quad \text{if}\quad x\in M-A.$$

Note that $A\cap (M-A)=\varnothing$, so $g$ is well-defined and $g$ is a bijection.

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